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Tensors

Anonymous Tue 13 Apr 21:52:58 2021 No.12971090 View Reply Original Report

Quoted By: >>12971336 >>12973562 >>12973582 >>12974087 >>12974103 >>12975636

Learning about tensors for a manifolds course, and we've been given a pretty reasonable definition of a tensor as an element of a tensor product, and mainly focusing on the exterior algebra.

However whenever I go online, pretty much every resource seems to focus on a physics/mapping/multidimensional array definition. Even the wikipedia article relegates the mathematical version to another page.

Why is the mathematics given such little importance?

Anonymous

Anonymous Tue 13 Apr 2021 22:44:32 No.12971336 Report

Quoted By: >>12971485 >>12973513 >>12973562 >>12973582

>>12971090
Is F a functor?

Anonymous

Anonymous Tue 13 Apr 2021 23:17:39 No.12971485 Report

Quoted By: >>12971517

>>12971336
According to my lecture notes, F(U,V) is the vector space generated by pairs (v,w) with v in V, w in W

Anonymous

Anonymous Tue 13 Apr 2021 23:24:43 No.12971517 Report

Quoted By:

>>12971485
i mean F(V,W)

Anonymous

Anonymous Wed 14 Apr 2021 12:07:54 No.12973513 Report

Quoted By: >>12973562 >>12973582

>>12971336
an operator that transform one vector into another vector

Anonymous

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Anonymous Wed 14 Apr 2021 12:19:12 No.12973562 Report

Quoted By:

>>12973513
It's acting on bigger sets than a vector

>>12971336
It's the free functor.

1. You take all pairs with components in V and W, say
(v, w)
2. You then pass to the (free) vector space spanned by all those, over some field K. So e.g.
k · (v, w)
suddenly makes sense, which is shorthand for
(k·v, k·w)
3. You then divide out an equivalence relation to that
e.g. the element
(c·v, w)
equals
(v, c·w)

>>12971090
Because the math is messy. More specifically, equivalence classes have uncanny rank and you need strong axioms to claim they are sets.
Consider for example that the real number 7 is an equivalence class (of sequences from the naturals to the rationals) of u. That is to say, the cardinality of this 7 is that of the reals
$|7|=|{\mathbb R}|$ .

Anonymous

Anonymous Wed 14 Apr 2021 12:20:26 No.12973570 Report

Quoted By:

so that

Anonymous

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Anonymous Wed 14 Apr 2021 12:23:09 No.12973582 Report

Quoted By: >>12973675

>>12973513
It's acting on bigger sets than a vector

>>12971336
It's the free functor.

1. You take all pairs with components in V and W, say
(v, w)

2. You then pass to the (free) vector space spanned by all those, over some field K. So e.g. multiplication
k · (v, w)
makes sense, which is shorthand for
(k·v, k·w)

3. You then divide out an equivalence relation so that
e.g. the element
(c·v, w)
equals
(v, c·w)
i.e. those are then not different elements anymore.

>>12971090 (OP)
Because the math is messy. More specifically, equivalence classes have uncanny rank and you need strong axioms to claim they are sets.
Consider for example that the real number 7 is an uncountable equivalence class of sequences from the naturals to the rationals. That is to say, the cardinality of this 7 is that of the reals
|7|=|R|.

Anonymous

Anonymous Wed 14 Apr 2021 12:44:05 No.12973675 Report

Quoted By: >>12973726

>>12973582
>More specifically, equivalence classes have uncanny rank and you need strong axioms to claim they are sets.
>Consider for example that the real number 7 is an uncountable equivalence class of sequences from the naturals to the rationals. That is to say, the cardinality of this 7 is that of the reals |7|=|R|.
you're right, *this* is the reason why we dumb down tensors for engineers. jesus.

Anonymous

Anonymous Wed 14 Apr 2021 12:56:47 No.12973726 Report

Quoted By:

>>12973675
Not sure if that's irony or not, but I'd say yes.
All those things which are gigantic per definition but in usage are characterized by a few simple computational rules, such as
>"you can do (20, 3) = (2, 30)" for elements in a tensor product of vector spaces but not for plain ordered pairs of vectors
The formal definition given by teachers who are also just learned engineer types instills just (arguably unnecessary, for their use) confusion in the students.
So those objects just get introduced operationally.

Anonymous

Anonymous Wed 14 Apr 2021 12:59:01 No.12973733 Report

Quoted By:

(and in the case of tensors, the situation is even more complicated since the object can be, and is, often characterized in pure arrow theoretical terms - although this only helps if everything you want to prove with them is equally just arrow theoretical)

Anonymous

Anonymous Wed 14 Apr 2021 13:48:39 No.12973878 Report

Quoted By: >>12973882

things you need to know to grasp tensors:

1. instill into your mind that a vector $v \in V$ with coordinates $v^i$ in whatever basis is NOT a column $(v^1,\dots,v^n)^T$ but the linear combination $a^1e_1 + \dots + a^n e_n$ where $(e_i)$ is the basis
2. the tensor product as an operation obeys $(au+bv) \otimes w = a\cdot u\otimes w + b\cdot v\otimes w$ , similarly in the second variable
3. if $(e_i)$ is a basis of $V$ , then $(e_i \otimes e_j)$ is a basis of $V \otimes V$

now it's just plug and chug.
if $u = u^ie_i$ and $v = v^j e_j$ , then $u\otimes v = u^iv^j \cdot e_i \otimes e_j$ , this is your "outer product".
if you have a different basis $(\tilde{e_j})$ related to the original by $e_i = a_i^j \tilde{e_j}$ and a tensor $T = T^{ij}e_i \otimes e_j$ , you just plug in the new basis to get $T = T^{ij} \cdot (a_i^k \tilde{e_k} \otimes a_j^l \tilde{e_l}) = T^{ij}a_i^ka_i^k \cdot \tilde{e_k}\otimes\tilde{e_l}$ . this is your "transformation law".

no need to memorize any formulas, derive everything on the spot. there you go, tensors explained in 5 minutes (and 100% mathematically correct).

Anonymous

Anonymous Wed 14 Apr 2021 13:49:49 No.12973882 Report

Quoted By:

>>12973878
> $T = T^{ij} \cdot (a_i^k \tilde{e_k} \otimes a_j^l \tilde{e_l}) = T^{ij}a_i^ka_i^k \cdot \tilde{e_k}\otimes\tilde{e_l}$
works in preview

Anonymous

Anonymous Wed 14 Apr 2021 14:04:10 No.12973913 Report

Quoted By:

Tensors are just multiplication. I'm not sure why categorists have to make it so difficult.

Anonymous

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Anonymous Wed 14 Apr 2021 14:38:09 No.12974012 Report

Quoted By: >>12974033 >>12974178

>engineering student: I'm having trouble with tensors, help pls
>math student: sure, it's super easy actually, so you have a bilinear map from a cartesian product (you know what a bilinear map is, right?) but, you see, the thing is that you don't like that it's only bilinear, you want to make it linear, but you also want to do it a nice way, so what you do is that you take all possible linear combinations of vectors in the product and you declare that some of them are actually the same thing..
>engineering student: sorry, I was wondering if you could explain the transformation rules
>math student: oh you mean THAT, well I don't work in explicit bases, sorry pal.

Anonymous

Anonymous Wed 14 Apr 2021 14:45:39 No.12974033 Report

Quoted By:

>>12974012
The red pill is once you get a basic understanding of the autistic math level the applied part becomes easy

Anonymous

Anonymous Wed 14 Apr 2021 14:59:30 No.12974087 Report

Quoted By: >>12974095 >>12974793 >>12975171

>>12971090
>reasonable definition of a tensor as an element of a tensor product
Are you not using the real/complex valued multi-linear function definition? That's how you end up applying tensors in general in most fields in general.
>However whenever I go online, pretty much every resource seems to focus on a physics/mapping/multidimensional array definition. Even the wikipedia article relegates the mathematical version to another page.
Here is the real and most general definition:
Let $R$ be a ring. Let $A_R$ be a right $R$ -module and ${}_R B$ be a left $R$ -module. Consider the additive Abelian group $F$ generated by $A_R \times {}_R B$ . Let $K$ be the subgroup of $A_R \times {}_R B$ generated by elements of the form $(a_1,b) + (a_2, b) - (a_1 + a_2, b)$ , $(a,b_1) + (a, b_2) - (a, b_1+b_2)$ , $(a r, b) - (a, r b)$ . We denote $A \otimes B = F/K$ to be the quotient group. An element of the form $a \otimes b \equiv (a,b) \mod K$ is called a tensor and $A\otimes B$ is called the tensor space. Note, that $A_R$ and ${}_R B$ are required to be a left or right module respectively for some other rings $S, T$ (or even $R$ ).

Anonymous

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Anonymous Wed 14 Apr 2021 15:02:14 No.12974095 Report

Quoted By: >>12974164

>>12974087
>thank you anon, it's all so clear now

Anonymous

Anonymous Wed 14 Apr 2021 15:04:23 No.12974103 Report

Quoted By:

>>12971090
I fucking hate you autists.
You can just express this as a phasor with two arguments. Just go to polar or cartesian for whatever operation you need to do.

Anonymous

Anonymous Wed 14 Apr 2021 15:24:41 No.12974164 Report

Quoted By:

>>12974095
You're welcome anon.

Anonymous

Anonymous Wed 14 Apr 2021 15:27:09 No.12974178 Report

Quoted By: >>12974589

>>12974012
>>engineering student: sorry, I was wondering if you could explain the transformation rules
What do you mean by transformation rules?

Anonymous

Anonymous Wed 14 Apr 2021 16:47:03 No.12974589 Report

Quoted By: >>12974840

>>12974178
the relation between tensor's coordinates in one basis and its coordinates in a different basis

Anonymous

Anonymous Wed 14 Apr 2021 17:13:49 No.12974793 Report

Quoted By:

>>12974087
Based.

Anonymous

Anonymous Wed 14 Apr 2021 17:27:41 No.12974840 Report

Quoted By: >>12974961 >>12977793

>>12974589
If $T: V\to W$ is a linear transformation, the induced linear transformation $V\otimes V \to W\otimes W$ is defined on decomposible elements of $V\otimes V$ by $T(v_1 \otimes v_2)] = T(v_1) \otimes T(v_2)$ . For a "change of coordinates" let $T be invertible$ and $W=V$ .

Anonymous

Anonymous Wed 14 Apr 2021 17:53:10 No.12974961 Report

Quoted By: >>12974982

>>12974840
looking at a change of coordinates as a map $T \colon V \to V$ is misleading desu

Anonymous

Anonymous Wed 14 Apr 2021 18:01:31 No.12974982 Report

Quoted By: >>12974986 >>12975041 >>12977793

>>12974961
>looking at a change of coordinates as a map T:VV is misleading desu
I also said $T: V \to V$ is invertible. It's not misleading. It's exactly what is happening. All invertible maps are a change of coordinates on a basis $e_1, ..., e_n, ...$ defined by $f_i = T(e_i)$ . All changes of basis are of that form for a particular map $T\in GL(V)$ . If you're talking about the change of an orthonormal basis given a symmetric positive definite form $Q: V\times V \to \re$ , then $T\in O(V,Q)\{ T\in GL(V) : Q(Tv, Tw) = Q(v,w)\; \forall v,w\in V \}$ .

Anonymous

Anonymous Wed 14 Apr 2021 18:02:38 No.12974986 Report

Quoted By:

>>12974982
\re should be $\mathbb{R}$ .

Anonymous

Anonymous Wed 14 Apr 2021 18:14:49 No.12975041 Report

Quoted By: >>12975179

>>12974982
change of coordinates means (informally) that the objects stay the same, you only write them differently. nothing is happening. the fact that every two bases differs by an invertible transformation is true and not that much relevant.

Anonymous

Anonymous Wed 14 Apr 2021 18:44:55 No.12975171 Report

Quoted By: >>12978061

>>12974087
OP here, we only learnt them for differential forms and the wedge product, so the only properties we really needed to know were antisymmetry, and by extension v^v = 0

Anonymous

Anonymous Wed 14 Apr 2021 18:45:50 No.12975179 Report

Quoted By: >>12975232 >>12977793

>>12975041
>change of coordinates means (informally) that the objects stay the same
>informally
>stay the same
After defining a basis through an invertible function, you can ask what the coordinates of a vector under the change of basis formula. The transformation on the vector space induces a transformation on the coordinate space given that if $e_1,..., e_n$ is a basis and $(c_1,...,c_n)$ is the coordinate of the element $v=c^ie_i$ there is an induced invertible transformation on the space of coordinates $S: F^n \to F^n$ where $F$ is the field over which $V$ is defined given the invertible transformation $T: V\to V$ defined by $(f_1\; ... \; f_n)(k_1,...,k_n) = (Te_1\; ... \; Te_n ) S(T)(c_1,...,c_n)$ where $k_1 = S(T)(c_1,...,c_n)$ . $S(T) \in GL(F^n)$ and $S(T)$ is written since it is uniquely determined by a transformation $T\in GL(V)$ given a basis. The coordinates do not remain the same and the basis does not remain the same. The study of a change of basis is always the study of invertible automorphisms on a given vector space since the change of basis is determined by an element of $GL(V)$ . The study of a change of bases is reducable to a study of $GL(V)$ and is derived from the study of $GL(V)$ . Thus the study of change of coordinates/change of basis is the study of $GL(V)$ .
>you only write them differently. nothing is happening.
No. Something is happening. See above
>the fact that every two bases differs by an invertible transformation is true
Yes.
>and not that much relevant.
No. See above.

Anonymous

Anonymous Wed 14 Apr 2021 18:57:07 No.12975232 Report

Quoted By:

>>12975179
in your eyes, changing a basis in V means, conceptually, that V and its elements actually undergo a linear transformation?

Anonymous

Anonymous Wed 14 Apr 2021 20:41:41 No.12975636 Report

Quoted By:

>>12971090
The real reason is because it is a more fundamental object than other objects you are familiar with. You don't need to understand what the tensor product is, you need understand how other operations are special cases of this object. There is no point in teaching you a semester's worth of group theory and finite vector spaces when a much simpler framework is "there's this operation called a tensor and this is what it does". If you want to understand the math part take those courses in your own time.

Applied math teaches constructivism because the way all this shit is used in real life does not depend on thousands of pages of elegant theorems and proofs, but rather on just fucking using it already.

Anonymous

Anonymous Thu 15 Apr 2021 08:19:39 No.12977793 Report

Quoted By:

>>12975179
>>12974982
>>12974840
a "basis" or "coordinates" is an isomorphism $\alpha \colon V \to \mathbb{F}^n$ . a "change of basis" from $\alpha$ to $\beta$ is the commutative triangle with edges $\alpha \colon V \to \mathbb{F}^n$ , $\beta \colon V \to \mathbb{F}^n$ and $\beta \circ \alpha^{-1} \colon \mathbb{F}^n \to \mathbb{F}^n$ , the last being the "transition matrix". there's literally no reason to invoke endomorphisms of $V$ , it just adds to confusion.

Anonymous

Anonymous Thu 15 Apr 2021 10:18:56 No.12978061 Report

Quoted By:

>>12975171
The important fact that e.g.
(5dx) ? (6dy) = (10dx) ? (3dy)
etc., doesn't follow from that nor does it hold for vector components

Anonymous

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Anonymous Thu 15 Apr 2021 17:15:07 No.12979485 Report

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